Question

How do you solve $x\left( x-7 \right)=0$?

Answer 1

Hint: The given polynomial is already in its factored form. It’s the multiplied form of two polynomials. Multiplication of two polynomials giving value of 0. Then at least one of them has to be 0. From this condition we find the values of the roots of the polynomial $x\left( x-7 \right)=0$. We also have the quadratic solving method for equation $a{{x}^{2}}+bx+c=0$. We put the root values to verify the result.

Complete step-by-step solution:
The given polynomial $x\left( x-7 \right)=0$ is already in its factored form. The multiplied terms cannot be further factorised.
Therefore, $x\left( x-7 \right)=0$ has multiplication of two polynomials giving value of 0. This means at least one of them has to be 0.
This gives either $x=0$ or $\left( x-7 \right)=0$. Simplifying we get
So, values of x are $x=0,7$.
We now multiply the terms and get the quadratic equation as $x\left( x-7 \right)=0$.
${{x}^{2}}-7x=0$
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have ${{x}^{2}}-7x=0$. The values of a, b, c is $1,-7,0$ respectively.
We put the values and get x as \[x=\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\times 1\times 0}}{2\times 1}=\dfrac{7\pm 7}{2}=0,7\].

Note: We find the value of x for which the function $f\left( x \right)=x\left( x-7 \right)$. We can see $f\left( 0 \right)=0\left( 0-7 \right)=0$. So, the root of the $f\left( x \right)=x\left( x-7 \right)=0$ will be the function $x$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. We can also do the same process for $\left( x-7 \right)$. We can see $f\left( 7 \right)=7\left( 7-7 \right)=7\times 0=0$. So, the root of the $f\left( x \right)=x\left( x-7 \right)=0$ will be the function $\left( x-7 \right)$.